Let the image of the point P(1, 2, 3) in the line L : (x-6)/3 = (y-1)/2 = (z-2)/3 be Q. let R(α, β,

Q: Wait, why do we need to find the foot of the perpendicular $M$ if the question asks about $Q$?

That is a brilliant question! While the question asks about $Q$, finding $M$ is the most direct path. Since $M$ is the midpoint of $PQ$, knowing $M$ gives you the geometric anchor for the entire segment. It is much easier to solve for the foot of the perpendicular than to jump straight to the image $Q$.

Q: Can I use the formula for the image of a point directly?

You could, but in JEE Advanced, memorizing formulas can be a trap. Deriving the result using the foot of the perpendicular $M$ ensures you understand the underlying geometry. Plus, as we saw in this problem, the shortcut $R$ being the midpoint of $PM$ only becomes obvious if you have already found $M$!

Q: Is there a faster way to find $R$ without calculating $Q$?

Absolutely! That is the beauty of this problem. Since $M$ is the midpoint of $PQ$, we know $PM = \frac{1}{2}PQ$. The problem tells us $R$ divides $PQ$ in a $1:3$ ratio, meaning $PR = \frac{1}{4}PQ$. By comparing these, we see $PR = \frac{1}{2}PM$, which means $R$ is simply the midpoint of $PM$. No need to find $Q$ at all!

Question

Let the image of the point $P(1, 2, 3)$ in the line $L : \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}$ be $Q$. let $R(\alpha, \beta, \gamma)$ be a point that divides internally the line segment $PQ$ in the ratio $1 : 3$. Then the value of $22(\alpha + \beta + \gamma)$ is equal to ____.

Sigma Percentile · Accepted Answer

### The Geometry of Reflection: A 3D Journey

Welcome, fellow explorer of the mathematical universe! Today, we are not just solving a problem; we are embarking on a journey through 3D space.

We are given a point $P(1, 2, 3)$ and a line $L$ defined by the symmetric equations:

$$\frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}$$

Our mission is to find a point $R$ that divides the segment $PQ$—where $Q$ is the image of $P$ in $L$—in a $1:3$ ratio. It sounds daunting, but let us break it down into a story of vectors and symmetry.

### Phase 1: The Foot of the Perpendicular

Before we can even think about the image $Q$, we must find the 'foot of the perpendicular' from $P$ to the line $L$. Let us call this point $M$. Think of $M$ as the point on the line $L$ that is closest to $P$.

Because $M$ lies on the line $L$, we can represent its coordinates using a single parameter, $\lambda$. By setting:

$$\frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3} = \lambda$$

We get the general coordinates of any point on the line as $M = (3\lambda + 6, 2\lambda + 1, 3\lambda + 2)$. This is our gateway to the solution.

### Phase 2: The Perpendicularity Condition

Now, we need to ensure that the line segment $PM$ is perpendicular to the line $L$. In the language of vectors, this means the dot product of the vector $\vec{PM}$ and the direction vector of the line $L$ must be zero.

The direction vector of $L$ is simply the denominator of our symmetric equation: $\vec{v} = (3, 2, 3)$. The vector $\vec{PM}$ is found by subtracting the coordinates of $P$ from $M$:

$$\vec{PM} = (3\lambda + 5, 2\lambda - 1, 3\lambda - 1)$$

Setting the dot product to zero, we get the equation:

$$3(3\lambda + 5) + 2(2\lambda - 1) + 3(3\lambda - 1) = 0$$

Expanding this, we find $9\lambda + 15 + 4\lambda - 2 + 9\lambda - 3 = 0$, which simplifies beautifully to $22\lambda + 10 = 0$. Thus, we find:

$$\lambda = -\frac{5}{11}$$

### Phase 3: The Geometric Shortcut

Here is where the magic happens. Many students would rush to find the coordinates of $Q$ using the midpoint formula. But wait! If $Q$ is the image of $P$, then $M$ is the midpoint of $PQ$.

The problem asks for a point $R$ that divides $PQ$ in a $1:3$ ratio. This means $PR = \frac{1}{4}PQ$. Since $PM = \frac{1}{2}PQ$, it follows that $PR = \frac{1}{2}PM$.

This reveals a stunning shortcut: **$R$ is simply the midpoint of $PM$!** We do not need to calculate $Q$ at all.

### Phase 4: The Final Calculation

With $\lambda = -\frac{5}{11}$, we find the coordinates of $M$:

$$M = \left( 3\left(-\frac{5}{11}ight) + 6, 2\left(-\frac{5}{11}ight) + 1, 3\left(-\frac{5}{11}ight) + 2 ight) = \left( \frac{51}{11}, \frac{1}{11}, \frac{7}{11} ight)$$

Now, we find $R$ by taking the midpoint of $P(1, 2, 3)$ and $M\left(\frac{51}{11}, \frac{1}{11}, \frac{7}{11}ight)$. The coordinates of $R(\alpha, \beta, \gamma)$ are:

$$\alpha = \frac{1 + \frac{51}{11}}{2} = \frac{31}{11}, \quad \beta = \frac{2 + \frac{1}{11}}{2} = \frac{23}{22}, \quad \gamma = \frac{3 + \frac{7}{11}}{2} = \frac{20}{11}$$

Finally, we sum these:

$$\alpha + \beta + \gamma = \frac{31}{11} + \frac{23}{22} + \frac{20}{11} = \frac{62 + 23 + 40}{22} = \frac{125}{22}$$

The question asks for $22(\alpha + \beta + \gamma)$, which is:

$$22 	imes \frac{125}{22} = \mathbf{125}$$

Isn't it elegant? By understanding the geometry, we bypassed the tedious calculation of $Q$ and arrived at the answer with clarity and confidence.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$G e n er a l P o in t M o n L in e L$

$D i r ec t i o n R a t i os o f P M$

$P er p e n d i c u l a r i t y C o n d i t i o n$

$S o l v in g f or λ$

$C oor d ina t es o f F oo t M$

$G eo m e t r i c R e l a t i o n s hi p o f R$

$C oor d ina t es o f R (α, β, γ)$

$S u m o f C oor d ina t es$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Equation of a Line in Space

The Geometry of Reflection

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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Consider a line $L$ passing through the points $P (1, 2, 1)$ and $Q (2, 1, - 1)$ . If the mirror image of the point $A (2, 2, 2)$ in the line $L$ is $(α, β, γ)$ , then $α + β + 6 γ$ is equal to

Let $(α, β, γ)$ be the image of the point $(8, 5, 7)$ in the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{5}$ . Then $α + β + γ$ is equal to :

If the mirror image of the point $P (3, 4, 9)$ in the line $\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1}$ is $(α, β, γ)$ , then $14 (α + β + γ)$ is :

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } General Point M on Line L

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Direction Ratios of PM

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Perpendicularity Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for λ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coordinates of Foot M

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Geometric Relationship of R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coordinates of R(α,β,γ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sum of Coordinates

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation